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1 year ago

14

At a bakery, Victor makes 20 cupcakes. He puts themin equal rows with 10 cupcakes in each row. Whichcalculation can be used to f

ind the number of rows ofcupcakes Victor makes?

1 answer:

Anna [14]1 year ago

4 0

Calculation = 20/10

Number of cupcakes Victor makes = 20

He puts them in equal rows:

Each row contains = 10 cupcakes

To know the number of rows he placed the 20 cupcakes,

we would divide 20 by 10

$\begin{gathered} \text{Number of rows = }\frac{20}{10} \\ \text{number of rows = 2} \end{gathered}$

Calculation = 20/10

The number of rows of cupcakes Victor makes = 2

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A new shopping mall records 120 total shoppers on their first day of business. Each day after that, the number of shoppers is 10

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Answer:

1138

Step-by-step explanation:

From the information given:

We can represent it perfectly in an exponential form:

$m = p(q)^x$

where;

p = initial value = 120

q = base of the exponential form

q = 1 + r

here; r = rate in decimal = 10% = 0.1

Then q can now be = 1 + 0.1 = 1.1

Replacing it into the exponential form, we get:

$m = 120(1.1)^x$

where;

x = number of days and m = number of shoppers

Thus:

For the first day:

$m = 120(1.1)^x$

m = 120

For the second day:

$m = 120(1.1)^1$

m = 132

For the third day:

$m = 120(1.1)^2$

m = 145.2

For the fourth day

$m = 120(1.1)^3$

m = 159.72

For the fifth-day

$m = 120(1.1)^4$

m = 175.692

For the sixth-day

$m = 120(1.1)^5$

m = 193.2612

For the seventh-day

$m = 120(1.1)^6$

m = 212.58732

Thus; the total numbers of shoppers for the first 7 days is:

$= 120+ 132 + 145.2 + 159.72 + 175.692+193.2612+212.58732$

= 1138.46052

≅ 1138

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2 years ago

Jacks average in math for the first nine weeks was an 88 . His second nine weeks average decreased 12.5% what was his average fo

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10% of 88 = 8.8

2.5% of 88 = 2.2

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Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

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Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

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$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$

Applying the ratio test:

$\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}$

$\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

Applying n → ∞

$\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

The numerator as well denominator of $\frac{a_{n}}{a_{n+1}}$ are polynomials of fifth degree with leading coefficients:

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2 years ago

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Answer:

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